how to collect maximum and minimum values from a grep search pattern using bash?

You were close, trying to fix OP's attempt here. This could be done in GNU grep like following. We could use GNU grep's -P option to enable PCRE regex there.

grep -oP '.*range: X_PER_Y_USER_RANGE_0\[1\]\([a-zA-Z]+, \d+\)=\K\d+'  Input_file

Explanation: Simple explanation would be, firstly enabling -oP options for GNU grep to enable PCRE regex capabilities and to get only matched values. Then in main program of grep, using regex .*range: X_PER_Y_USER_RANGE_0\[1\]\([a-zA-Z]+, \d+\)= to match everything from starting to till (), note here escaping part of [, ],(,) to make those characters as literal ones. Then using \K option to forget all the matched values till now. Mentioning \d+ will match digits after it which is OP's actual requirement to get the digits.

You just needed to escape the special regex characters :

grep -Po 'range: X_PER_Y_USER_RANGE_0\[1\]\(a, \d\)=\K\d+' file.txt
121
256879
2300
56897132
12579
968746

For such a case (to me) Perl itself is preferable :

perl -lne '/ \d\)=\K\d+/ && print $&' file.txt
121
256879
2300
56897132
12579
968746

and for min and max something like this:

perl -lne '/ \d\)=\K\d+/g && push(@number,$&); END{ print "@{[sort {$a <=> $b} @number]}[0,-1]"}' file.txt
121 56897132

and END { ... } means sort them and take out 0 and last index of the array of @number

For sort using bash and not Perl to have min and max you can pipe your output to:

... | { mapfile -t arr; paste <(sort -n <(tr ' ' '\n' <<<  ${arr[@]})) <(sort -rn <(tr ' ' '\n' <<<  ${arr[@]})); }
121 56897132
2300    968746
12579   256879
256879  12579
968746  2300
56897132    121

which creates an array and then sort it twice. Then just pipe it to head -n 1 e.g.

... | head -n 1
121 56897132